Question

For one mole of a Van der Waals' gas when b = 0 and T = 300 K, the pV vs 1/V plot is shown below. The value of the van der Waals' constant a (atm L $mo{l}^{-2}$)

• A. 1
• B. 4.5
• C. 1.5
• D. 3

Answer 1

The correct option is C 1.5

Van der Waals' equation of state is given as:
$(p+\frac{n^{2}a}{V^2})(V-nb)=nRT$

Given,
n = 1 and b = 0
$(p+\frac{a}{V^2})V=RT$
$\Rightarrow pV=RT-\frac{a}{V}$ ....(i)

This is an equation of straight line between pV and$\frac{1}{V}$ whose slope - a.

Slope of the given graph is,
$-a=\frac{20.1-21.6}{3-2}=-1.5$
$\Rightarrow$ a = 1.5